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Motion in A Straight Line question

2024 · 29 Jan · Shift 2 · Q74
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  5. /2024 · 29 Jan · Shift 2 · Q74

Motion in A Straight Line question

2024 · 29 Jan · Shift 2 · Q74

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
A particle is moving in a straight line. The variation of position 'xxx' as a function of time 'ttt' is given as x=(t3−6t2+20t+15)mx=\left(t^3-6 t^2+20 t+15\right) mx=(t3−6t2+20t+15)m. The velocity of the body when its acceleration becomes zero is :
  1. A
    6 m/s
  2. B
    10 m/s
  3. C
    8 m/s
  4. D
    4 m/s
View written solutionFree

Correct answer: C

  1. Given position function

x(t)=t3−6t2+20t+15x(t)=t^3-6t^2+20t+15x(t)=t3−6t2+20t+15

We need the velocity when acceleration becomes zero.

  1. Find velocity

Velocity is the first derivative of position:

v=dxdt=3t2−12t+20v=\frac{dx}{dt}=3t^2-12t+20v=dtdx​=3t2−12t+20

  1. Find acceleration

Acceleration is the derivative of velocity:

a=dvdt=6t−12a=\frac{dv}{dt}=6t-12a=dtdv​=6t−12

  1. Set acceleration to zero

6t−12=06t-12=06t−12=0

6t=126t=126t=12

t=2 st=2\text{ s}t=2 s

  1. Find velocity at t=2t=2t=2 s

Substitute into the velocity expression:

v(2)=3(2)2−12(2)+20v(2)=3(2)^2-12(2)+20v(2)=3(2)2−12(2)+20

v(2)=3⋅4−24+20v(2)=3\cdot 4-24+20v(2)=3⋅4−24+20

v(2)=12−24+20=8 m/sv(2)=12-24+20=8\text{ m/s}v(2)=12−24+20=8 m/s

  1. Match with options

8 m/s8\text{ m/s}8 m/s corresponds to Option C.

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